Problem: The power generated by an electrical circuit (in watts) as a function of its current $c$ (in amperes) is modeled by: $P(c)=-20(c-3)^2+180$ Which currents will produce no power (i.e. $0$ watts)? Enter the lower current first. Lower current:
Answer: The circuit's power is $0$ when $P(c)=0$. $\begin{aligned} P(c)&=0 \\\\ -20(c-3)^2+180&=0 \\\\ -20(c-3)^2&=-180 \\\\ (c-3)^2&=9 \\\\ \sqrt{(c-3)^2}&=\sqrt{9} \\\\ c-3&=\pm3 \\\\ c&=\pm3+3 \\\\ c=6&\text{ or }c=0 \end{aligned}$ In conclusion, these are the currents that will produce no power: Lower current: $0$ amperes Higher current: $6$ amperes